Calculate concentration changes on dilution using c1v1 = c2v2 and predict the effect of dilution on pH for strong and weak acid and base solutions

What this dot point is asking

NESA wants you to handle the quantitative side of acid-base chemistry: convert between concentration units, perform serial and one-step dilutions, and predict how pH changes when an acid or base is diluted. The mathematics is the dilution identity $c_1 V_1 = c_2 V_2$; the chemistry is the difference between strong and weak acids on dilution, which builds on the strong vs weak page and feeds into titration analysis.

The answer

Concentration units

Molarity is the default HSC unit because it links directly to stoichiometry ($n = cV$). The other units appear in environmental and biological contexts (lead in drinking water, residual chlorine, salinity).

Conversions to remember.

• $1 \text{ ppm} \approx 1 \text{ mg/L}$ for dilute aqueous solutions (because 1 L of water has mass 1 kg).
• $c \text{ (mol/L)} = \frac{\text{percent w/v} \times 10}{M_r}$, where $M_r$ is the molar mass in g/mol.

The dilution equation

When solvent is added to a solution, the moles of solute do not change. Therefore:

Use any consistent volume unit (mL or L) as long as both sides match. The equation works for any solute, not just acids and bases.

Procedure for an accurate dilution.

1. Calculate the volume of stock needed: $V_1 = c_2 V_2 / c_1$.
2. Transfer $V_1$ to a volumetric flask of size $V_2$ using a pipette (for small volumes) or a measuring cylinder.
3. Add distilled water to roughly three-quarters full, swirl.
4. Top up to the calibration mark, with a dropper for the last few drops.
5. Stopper, invert several times to ensure homogeneity.

pH on dilution: strong acid or base

For a strong acid, $[H^+] = c$ (fully ionised). On dilution, $[H^+]$ drops in direct proportion to $c$. A 10-fold dilution raises pH by 1.0 unit; a 100-fold dilution raises pH by 2.0 units.

For a strong base, $[OH^-] = c$ on the same logic, and pH falls by 1 unit per 10-fold dilution (because pOH rises by 1).

Limit: as you dilute past about $10^{-6}$ mol/L, the auto-ionisation of water becomes significant and pH approaches 7 from below (for an acid) or above (for a base), never crossing 7. A common HSC error is to claim that $10^{-9}$ mol/L $HCl$ has pH 9; in fact it has pH just under 7 because the dominant source of $H^+$ is now water itself.

pH on dilution: weak acid or base

For a weak acid, $[H^+] \approx \sqrt{K_a c}$ (from the ICE shortcut, valid when ionisation is small).

A 10-fold dilution changes $c$ by 10, so $[H^+]$ changes by $\sqrt{10} \approx 3.16$, and pH rises by $\log_{10}(\sqrt{10}) = 0.5$. Weak acids resist pH change on dilution more than strong acids do.

Mechanistically (Le Chatelier): the ionisation equilibrium has more moles of dissolved particles on the right side, so dilution favours further ionisation. The percent ionisation rises as concentration falls, partially offsetting the dilution.

Summary rule. A 10-fold dilution raises pH by:

• 1.0 unit for a strong acid (until water dominates).
• ~0.5 unit for a weak acid.
• ~0 for a buffer (until exhausted).

Serial dilution

If you need a very dilute solution, a single one-step dilution can be inaccurate (pipetting a very small volume). Serial dilution does it in stages: each step is a 10- or 100-fold dilution. To go from 1.00 mol/L to $1.00 \times 10^{-4}$ mol/L, do four successive 10-fold dilutions.

Examples in context

Example 1. Diluting bench acid in the NSW HSC prac lab. A standard exercise has students prepare 250 mL of 0.100 mol L$^{-1}$ HCl from a 2.00 mol L$^{-1}$ bench stock. Rearranging $c_1 V_1 = c_2 V_2$ gives $V_1 = 0.100 \times 0.250 / 2.00 = 12.5$ mL. The student pipettes 12.5 mL of stock into a 250 mL volumetric flask, swirls, and tops up to the mark. NSW DET safety guidelines require acid into water (never the reverse) and use of a safety pipette filler. The resulting solution has pH 1.00 because HCl is fully ionised. The same calculation underpins every clinical IV dilution at NSW Health.

Example 2. Dilution effects on weak versus strong acid in environmental sampling. A WaterNSW officer collects acidic mine drainage near Captains Flat at $[H^+] = 0.010$ mol L$^{-1}$ (pH 2.0) and a sample of dilute acetic acid runoff from a vineyard at 0.010 mol L$^{-1}$ acetic acid ($K_a = 1.8 \times 10^{-5}$, pH 3.4). On a 100-fold dilution to $1.0 \times 10^{-4}$ mol L$^{-1}$, the strong acid drops to pH 4.0 (a rise of 2 units) but the weak acid rises only to about pH 4.7. Officers use this contrast to confirm whether a low pH reading reflects strong mineral acidity or weak organic acid, an evidentiary distinction that affects how runoff is regulated.

Try this

Q1. State the dilution equation $c_1 V_1 = c_2 V_2$ and explain why moles of solute are conserved on dilution. [2 marks]

• Cue. Equation conserves moles ($n = cV$); adding solvent does not create or destroy solute particles.

Q2. Calculate the volume of concentrated 18.0 mol L$^{-1}$ $H_2SO_4$ required to prepare 500 mL of 0.250 mol L$^{-1}$ acid. [2 marks]

• Cue. $V_1 = c_2 V_2 / c_1 = (0.250 \times 0.500) / 18.0 = 6.94$ mL.

Q3. A 0.10 mol L$^{-1}$ solution of HCl and a 0.10 mol L$^{-1}$ solution of acetic acid (pKa = 4.74) are each diluted 100-fold. (a) Calculate the pH of each before dilution. (b) Calculate the pH of each after dilution. (c) Account for the difference in pH change. [2+2+2 marks]

• Cue. (a) HCl pH = 1.0; acetic pH = 2.87. (b) HCl pH = 3.0; acetic pH approx 3.87. (c) Strong acid: pH rises by 2 (full ionisation); weak acid: equilibrium shifts right, pH rises by only 1.